Total position where king can reach on a chessboard in exactly M moves | Set 2
Given the position of the king on an 8 X 8 chessboard, the task is to count the total number of squares that can be visited by the king in m moves. The position of the king is denoted using row and column number.
Note: The square which is currently acquired by the king is already visited and will be counted in the result.
Examples:
Input: r = 4, c = 4, m = 1
Output: 9
Input: r = 4, c = 4, m = 2
Output: 25
Approach: A king can move one square in any direction (i.e horizontally, vertically and diagonally). So, in one move king can visit its adjacent squares.
So, A square which is within m units distance (Considering 1 Square as 1 unit distance) from the king’s current position can be visited in m moves.
For all squares of the chessboard, check if a particular square is at m unit distance away or less from King’s current position.
- Increment count, if step 1 is true.
- Print the count
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of squares // that can be visited by king in m moves int countSquares( int r, int c, int m) { // To store the count of squares int squares = 0; // Check all squares of // the chessboard for ( int i = 1; i <= 8; i++) { for ( int j = 1; j <= 8; j++) { // Check if square (i, j) is // at a distance <= m units // from king's current position if (max( abs (i - r), abs (j - c)) <= m) squares++; } } // Return count of squares return squares; } // Driver code int main() { int r = 4, c = 4, m = 1; cout << countSquares(r, c, m) << endl; return 0; } |
Java
// Java implementation of the approach class GFG { // Function to return the count of squares // that can be visited by king in m moves static int countSquares( int r, int c, int m) { // To store the count of squares int squares = 0 ; // Check all squares of // the chessboard for ( int i = 1 ; i <= 8 ; i++) { for ( int j = 1 ; j <= 8 ; j++) { // Check if square (i, j) is // at a distance <= m units // from king's current position if (Math.max(Math.abs(i - r), Math.abs(j - c)) <= m) squares++; } } // Return count of squares return squares; } // Driver code public static void main(String[] args) { int r = 4 , c = 4 , m = 1 ; System.out.print(countSquares(r, c, m)); } } |
C#
// C# implementation of the approach using System; class GFG { // Function to return the count of squares // that can be visited by king in m moves static int countSquares( int r, int c, int m) { // To store the count of squares int squares = 0; // Check all squares of // the chessboard for ( int i = 1; i <= 8; i++) { for ( int j = 1; j <= 8; j++) { // Check if square (i, j) is // at a distance <= m units // from king's current position if (Math.Max(Math.Abs(i - r), Math.Abs(j - c)) <= m) squares++; } } // Return count of squares return squares; } // Driver code public static void Main() { int r = 4, c = 4, m = 1; Console.Write(countSquares(r, c, m)); } } |
Python3
# Python implementation of the approach # Function to return the count of squares # that can be visited by king in m moves def countSquares(r, c, m): # To store the count of squares squares = 0 # Check all squares of # the chessboard for i in range ( 1 , 9 ): for j in range ( 1 , 9 ): # Check if square (i, j) is # at a distance <= m units # from king's current position if ( max ( abs (i - r), abs (j - c)) < = m): squares = squares + 1 # Return count of squares return squares # Driver code r = 4 c = 4 m = 1 print (countSquares(r, c, m)); |
PHP
<?php // PHP implementation of the approach // Function to return the count of squares // that can be visited by king in m moves function countSquares( $r , $c , $m ) { // To store the count of squares $squares = 0; // Check all squares of // the chessboard for ( $i = 1; $i <= 8; $i ++) { for ( $j = 1; $j <= 8; $j ++) { // Check if square (i, j) is // at a distance <= m units // from king's current position if (max( abs ( $i - $r ), abs ( $j - $c )) <= $m ) $squares ++; } } // Return count of squares return $squares ; } // Driver code $r = 4; $c = 4; $m = 1; echo countSquares( $r , $c , $m ); // This code is contributed by Ryuga ?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the count of squares // that can be visited by king in m moves function countSquares(r, c, m) { // To store the count of squares let squares = 0; // Check all squares of // the chessboard for (let i = 1; i <= 8; i++) { for (let j = 1; j <= 8; j++) { // Check if square (i, j) is // at a distance <= m units // from king's current position if (Math.max(Math.abs(i - r), Math.abs(j - c)) <= m) squares++; } } // Return count of squares return squares; } // Driver Code let r = 4, c = 4, m = 1; document.write(countSquares(r, c, m)); </script> |
9
Time Complexity: O(1), since the loop runs for a total of 64 times, that is constant time only.
Auxiliary Space: O(1), since no extra space has been taken.